# L'H¶pital's Rule

(Redirected from L'Hopital's Rule)

## Discovered By

Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli.

## The Rule

For $\frac {0}{0}$ or $\frac {\infty}{\infty}$ case, the limit $$\lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)}$$ where $f'(x)$ and $g'(x)$ are the first derivatives of $f(x)$ and $g(x)$, respectively.

## Examples $\lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48}$ $\lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1}$

## Proof of l'H├┤pital's rule

A standard proof of l'H├┤pital's rule uses Cauchy's mean value theorem. l'H├┤pital's rule has many variations depending on whether c and L are finite or infinite, whether f and g converge to zero or infinity, and whether the limits are one-sided or two-sided.

### Zero over zero

Suppose that c and L are finite and f and g converge to zero.

First, define (or redefine) f(c) = 0 and g(c) = 0. This makes f and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c).

Since $\lim_{x\to c}\frac {f'(x)}{g'(x)}$ exists, there is an interval (c − ╬┤c + ╬┤) such that for all x in the interval, with the possible exception of x = c, both $f'(x)$ and $g'(x)$ exist and $g'(x)$ is not zero.

If x is in the interval (cc + ╬┤), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [cx] (and a similar statement holds for x in the interval (c − ╬┤c)). The mean value theorem implies that g(x) is not zero (since otherwise there would be a y in the interval (cx) with $g'(y)=0$). Cauchy's mean value theorem now implies that there is a point ╬Šx in (cx) such that $\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.$

If x approaches c, then ╬Šx approaches c (by the squeeze theorem ). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, it follows that $\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

### Infinity over infinity

Suppose that L is finite, c is positive infinity, and f and g converge to positive infinity.

For every ╬Ą > 0, there is an m such that $\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.$

The mean value theorem implies that if x > m, then g(x) ŌēĀ g(m) (since otherwise there would be a y in the interval (mx) with $g'(y)=0$). Cauchy's mean value theorem applied to the interval [mx] now implies that $\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.$

Since f converges to positive infinity, if x is large enough, then f(x) ŌēĀ f(m). Write $\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.$

Now,

\begin{align} & \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align} (Error compiling LaTeX. ! Package amsmath Error: \begin{align} allowed only in paragraph mode.)

For x sufficiently large, this is less than ╬Ą and therefore $\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.$*
• Note: Steps are missing.